# Convergence in compact Hausdorff space

You are told that $x_n\to x$ in the sup norm. This means that, given $\varepsilon>0$, there exists N such that $n\geqslant N\ \Rightarrow\ \|x_n-x\|_{\text{sup}} = \sup\{|x_n(t) - x(t)|:t\in X\} <\varepsilon$. Take $t=p_n$ to see that $n\geqslant N\ \Rightarrow\ |x_n(p_n) - x(p_n)|<\varepsilon$. Thus $|x_n(p_n) - x(p_n)|\to0$ as $n\to\infty$.